3. Longest Substring Without Repeating Characters

Description

Given a string s, find the length of the longest substring without repeating characters.

Example 1:

Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

Constraints:

0 <= s.length <= 5 * 10⁴
s consists of English letters, digits, symbols and spaces.

Solutions

Solution 1: Two pointers + Hash Table

Define a hash table to record the characters in the current window. Let $i$ and $j$ represent the start and end positions of the non-repeating substring, respectively. The length of the longest non-repeating substring is recorded by ans.

For each character $s[j]$ in the string s, we call it $c$. If $c$ exists in the window $s[i..j-1]$, we move $i$ to the right until $s[i..j-1]$ does not contain c. Then we add c to the hash table. At this time, the window $s[i..j]$ does not contain repeated elements, and we update the maximum value of ans.

Finally, return ans.

The time complexity is $O(n)$, where $n$ represents the length of the string s.

Two pointers algorithm template:

for (int i = 0, j = 0; i < n; ++i) {
    while (j < i && check(j, i)) {
        ++j;
    }
    // logic of specific problem
}

Python3JavaC++GoTypeScriptRustJavaScriptC#PHPSwiftNim

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        ss = set()
        ans = i = 0
        for j, c in enumerate(s):
            while c in ss:
                ss.remove(s[i])
                i += 1
            ss.add(c)
            ans = max(ans, j - i + 1)
        return ans

class Solution {
    public int lengthOfLongestSubstring(String s) {
        boolean[] ss = new boolean[128];
        int ans = 0;
        for (int i = 0, j = 0; j < s.length(); ++j) {
            char c = s.charAt(j);
            while (ss[c]) {
                ss[s.charAt(i++)] = false;
            }
            ss[c] = true;
            ans = Math.max(ans, j - i + 1);
        }
        return ans;
    }
}

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        bool ss[128]{};
        int ans = 0;
        for (int i = 0, j = 0; j < s.size(); ++j) {
            while (ss[s[j]]) {
                ss[s[i++]] = false;
            }
            ss[s[j]] = true;
            ans = max(ans, j - i + 1);
        }
        return ans;
    }
};

func lengthOfLongestSubstring(s string) (ans int) {
    ss := [128]bool{}
    for i, j := 0, 0; j < len(s); j++ {
        for ss[s[j]] {
            ss[s[i]] = false
            i++
        }
        ss[s[j]] = true
        ans = max(ans, j-i+1)
    }
    return
}

function lengthOfLongestSubstring(s: string): number {
    let ans = 0;
    const ss: Set<string> = new Set();
    for (let i = 0, j = 0; j < s.length; ++j) {
        while (ss.has(s[j])) {
            ss.delete(s[i++]);
        }
        ss.add(s[j]);
        ans = Math.max(ans, j - i + 1);
    }
    return ans;
}

use std::collections::HashSet;

impl Solution {
    pub fn length_of_longest_substring(s: String) -> i32 {
        let s = s.as_bytes();
        let mut ss = HashSet::new();
        let mut i = 0;
        s
            .iter()
            .map(|c| {
                while ss.contains(&c) {
                    ss.remove(&s[i]);
                    i += 1;
                }
                ss.insert(c);
                ss.len()
            })
            .max()
            .unwrap_or(0) as i32
    }
}

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function (s) {
    let ans = 0;
    const ss = new Set();
    for (let i = 0, j = 0; j < s.length; ++j) {
        while (ss.has(s[j])) {
            ss.delete(s[i++]);
        }
        ss.add(s[j]);
        ans = Math.max(ans, j - i + 1);
    }
    return ans;
};

public class Solution {
    public int LengthOfLongestSubstring(string s) {
        int ans = 0;
        var ss = new HashSet<char>();
        for (int i = 0, j = 0; j < s.Length; ++j) {
            while (ss.Contains(s[j])) {
                ss.Remove(s[i++]);
            }
            ss.Add(s[j]);
            ans = Math.Max(ans, j - i + 1);
        }
        return ans;
    }
}

class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function lengthOfLongestSubstring($s) {
        $ans = 0;
        $ss = [];
        for ($i = 0, $j = 0; $j < strlen($s); ++$j) {
            while (in_array($s[$j], $ss)) {
                unset($ss[array_search($s[$i++], $ss)]);
            }
            $ss[] = $s[$j];
            $ans = max($ans, $j - $i + 1);
        }
        return $ans;
    }
}

class Solution {
    func lengthOfLongestSubstring(_ s: String) -> Int {
        var map = [Character: Int]()
        var currentStartingIndex = 0
        var i = 0
        var maxLength = 0
        for char in s {
            if map[char] != nil {
                if map[char]! >= currentStartingIndex {
                    maxLength = max(maxLength, i - currentStartingIndex)
                    currentStartingIndex = map[char]! + 1
                }
            }
            map[char] = i
            i += 1
        }
        return max(maxLength, i - currentStartingIndex)
    }
}

proc lengthOfLongestSubstring(s: string): int =
  var
    i = 0
    j = 0
    res = 0
    literals: set[char] = {}

  while i < s.len:
    while s[i] in literals:
      if s[j] in literals:
        excl(literals, s[j])
      j += 1
    literals.incl(s[i]) # Uniform Function Call Syntax f(x) = x.f
    res = max(res, i - j + 1)
    i += 1

  result = res # result has the default return value

3. Longest Substring Without Repeating Characters

Description

Solutions

Solution 1: Two pointers + Hash Table

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